A particle moves along the x axis according to the equation x = 50t + 10t2, wher

Question

A particle moves along the x axis according to the equation x = 50t + 10t2, where x is in meters and t is in seconds. Calculate the average velocity of the particle during the first 3.0 s of its motion. m/s Calculate the instantaneous velocity of the particle at t = 3.0 s. m/s Calculate the instantaneous acceleration of the particle at t = 3.0 s m/s2 Using Figure A, in which the blue parabola is the plot of x versus t, indicate how the average velocity during the first 3.0 s can be obtained graphically. It is the slope of the green line. It is the x-coordinate of the intersection of the red line and the blue graph. None of these It is the slope of the red line. It is the y-intercept of the purple line. Using the graph of x versus t (Figure A) where the blue parabola is the actual plot, indicate how the instantaneous velocity at t = 3.0 s can be obtained graphically. It is the intersection of the red and purple lines. None of these It is the slope of the purple line. It is the slope of the green line. It is the slope of the red line. From the three lines in Figure B, choose the correct plot of v versus t, and indicate how to find the acceleration at t = 3.0 s graphically. None of these answers are correct. The green line is the velocity graph and the x-coordinate of its bend is the acceleration. The blue line is the velocity graph and the y-coordinate of its intersection with the red/position graph is the acceleration. The blue line is the velocity graph and its slope is the acceleration. The green line is the velocity and the slope of the line connecting its endpoints is the acceleration.

Explanation / Answer

a) average velocity = x(3) -x(0)/3 = (50*3+10*3^2)/3=80 m/s b) instanteous v = dx/dt = 50+20t = 50+20*3 = 110 m/s c) a = dv/dt = 20 m/s^2 d) It is slope of red line e)slope of green line f)the blue line is the velocity graph and its slope is acceleration

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